Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, q;
vector<int> l, r, x, y;
vector<long long> sum;
double ans(int x1, int y1, int x, int y) {
  long long a = y1 - y;
  long long b = x - x1;
  long long c = -a * x - b * y;
  double xs = -1.0 * c / a;
  int ll = 0, rr = n - 1;
  while (ll <= rr) {
    int md = (ll + rr) / 2;
    if (r[md] < xs)
      ll = md + 1;
    else
      rr = md - 1;
  }
  if (ll == n) return sum[n - 1];
  double ans = 0;
  if (ll != 0) ans = sum[ll - 1];
  if (l[ll] < xs) ans += xs - l[ll];
  return ans;
}
int main() {
  int y1, a, b;
  cin >> y1 >> a >> b;
  cin >> n;
  l.resize(n);
  r.resize(n);
  for (int i = 0; i < n; ++i) cin >> l[i] >> r[i];
  cin >> q;
  x.resize(q);
  y.resize(q);
  for (int i = 0; i < q; ++i) cin >> x[i] >> y[i];
  sum.resize(n);
  sum[0] = r[0] - l[0];
  for (int i = 1; i < n; ++i) sum[i] = sum[i - 1] + r[i] - l[i];
  for (int i = 0; i < q; ++i) {
    cout << fixed << setprecision(10)
         << (ans(b, y1, x[i], y[i]) - ans(a, y1, x[i], y[i])) * (y[i] - y1) /
                y[i]
         << '\n';
  }
  return 0;
}